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Question: The value of \[{}^2{P_1} + {}^3{P_1} + ... + {}^n{P_1}\] is equal to A.\[\dfrac{{{n^2} - n + 2}}{2...

The value of 2P1+3P1+...+nP1{}^2{P_1} + {}^3{P_1} + ... + {}^n{P_1} is equal to
A.n2−n+22\dfrac{{{n^2} - n + 2}}{2}
B.n2+n+22\dfrac{{{n^2} + n + 2}}{2}
C.n2+n−12\dfrac{{{n^2} + n - 1}}{2}
D.n2−n−12\dfrac{{{n^2} - n - 1}}{2}
E.n2+n−22\dfrac{{{n^2} + n - 2}}{2}

Explanation

Solution

Here, we will use formula of permutation, nPr=n!(n−r)!{}^n{P_r} = \dfrac{{n!}}{{\left( {n - r} \right)!}}, where nn is the number of items, and rr represents the number of items being chosen. Then we will simplify it using 1+2+3+...+n=n(n+1)21 + 2 + 3 + ... + n = \dfrac{{n\left( {n + 1} \right)}}{2}.

Complete step-by-step answer:
We are given that 2P1+3P1+...+nP1{}^2{P_1} + {}^3{P_1} + ... + {}^n{P_1}.
We know that the formula of permutation, nPr=n!(n−r)!{}^n{P_r} = \dfrac{{n!}}{{\left( {n - r} \right)!}}, where nn is the number of items, and rr represents the number of items being chosen.
Using the above formula in the terms of the given series, we get

⇒2!(2−1)!+3!(3−1)!+...+n!(n−1)! ⇒2!1!+3!2!+...+n!(n−1)!  \Rightarrow \dfrac{{2!}}{{\left( {2 - 1} \right)!}} + \dfrac{{3!}}{{\left( {3 - 1} \right)!}} + ... + \dfrac{{n!}}{{\left( {n - 1} \right)!}} \\\ \Rightarrow \dfrac{{2!}}{{1!}} + \dfrac{{3!}}{{2!}} + ... + \dfrac{{n!}}{{\left( {n - 1} \right)!}} \\\

Simplifying the factorials in the above expression, we get
⇒2×1!1!+3×2!2!+...+n×(n−1)!(n−1)!\Rightarrow \dfrac{{2 \times 1!}}{{1!}} + \dfrac{{3 \times 2!}}{{2!}} + ... + \dfrac{{n \times \left( {n - 1} \right)!}}{{\left( {n - 1} \right)!}}
We will now cancel the same factorials in numerators and denominators in the above expression, we get
⇒2+3+...+n\Rightarrow 2 + 3 + ... + n
Adding and subtracting the above equation with 1, we get
⇒1+2+3+...+n−1\Rightarrow 1 + 2 + 3 + ... + n - 1
Using the formula, 1+2+3+...+n=n(n+1)21 + 2 + 3 + ... + n = \dfrac{{n\left( {n + 1} \right)}}{2} in the above equation and simplifying, we get

⇒n(n+1)2−1 ⇒n(n+1)−22 ⇒n2+n−22  \Rightarrow \dfrac{{n\left( {n + 1} \right)}}{2} - 1 \\\ \Rightarrow \dfrac{{n\left( {n + 1} \right) - 2}}{2} \\\ \Rightarrow \dfrac{{{n^2} + n - 2}}{2} \\\

Hence, option E is correct.

Note: In solving these types of questions, you should be familiar with the formula of permutations. Some students use the formula of combinations, nCr= n ‾!∣ r ‾!∣⋅ n−r ‾!∣{}^n{C_r} = \dfrac{{\left. {\underline {\, n \,}}\\! \right| }}{{\left. {\underline {\, r \,}}\\! \right| \cdot \left. {\underline {\, {n - r} \,}}\\! \right| }} instead of permutations is nPr=n!(n−r)!{}^n{P_r} = \dfrac{{n!}}{{\left( {n - r} \right)!}}, where nn is the number of items, and rr represents the number of items being chosen, which is wrong.